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How can I estimate Roll from Position?
I have flight test data where the IMU is junk. The roll and yaw are obviously wrong, while the pitch seems to be correct. [I know the yaw is wrong because it doesn't match the changes in headings, and I know the roll is wrong because the it stays near 0° during horizontal loops where it should be closer to 20°.] The pitch seems ok because it is stays near 5° most of the time, which was the angle of attack for most of the flight.
I also have GPS data that gives me position at a high update rate.
I was able to use the position to compute the yaw values. With the position updates I can also estimate speed. With the position I can get a turn radius, and the change in position gives a velocity which can lead to a linear and/or angular momentum.
Now, I'd like to get an estimate of the roll (I say estimate because I figure the computation will require some simplifying assumptions).
The resolution for roll that I need is much less than the position data (by a factor of about 1000).
My question is this: given the position of the aircraft, how can I estimate the roll?
The purpose of this is for analysis of flight test data for a product that was tested on the aircraft. The product had (what turned out to be) a poor IMU, and we cannot do another flight test. I need the attitude and position to complete my analysis.

Does this IMU only have an accelerometer?

@Collin  Yes, the IMU data has an accelerometer (3 values for each time step).

You can, but the accuracy is going to be somewhat poor.
Given the positions you can approximate the path with your favourite line approximation method. The important thing is that it can give you continuous evaluations (no jumps and no discontinuities) of at least the first and second derivative. Third order algebraic splines are a good approach.
Note that you will need to map Latitude and Longitude to some Cartesian representation. Depending on the length of the flight, a "flat earth" approximation might or might not be a good idea.
With these, assuming that each manoeuvre was mostly flat(*), you can compute your local curvature:
$$k = \frac{x'y'' - y'x''}{\sqrt{(x'^2 + y'^2)^3}}$$
Now you can estimate your bank angle $\phi$ with
$$\phi = atan\left( \frac{k \cdot V_{tas}^2}{g} \right)$$
where $g$ is Earth's gravity.

(*) the formula is for a manoeuvre in the x-y plane. You could turn your coordinates so to have a local x-y plane that is coincident with the plane where the aircraft was manoeuvering, but then you could not use the bank angle formula, that is correct only for turns in the horizontal plane (thanks Jan Hudec for correcting)

Ad turning your coordinates: no, you can't (easily), because the formula is only applicable for plane orthogonal to gravity. It's easier to use the full 3D formula.

@Frederico I think this is exactly what I need. The maneuvers are indeed (approximately) at constant altitude. I'll have to crunch the numbers to see if this gives reasonable values, but I like this approach because I don't have to define a turn radius (which would be difficult because planes don't necessarily turn in perfect circles with a constant radius).

@JanHudec corrected, thanks.

@Baller (it's FEderico) please note that you are defining a turn radius anyway, $k=1/r$, I am simply giving you a straightforward way of defining it

@Federico I ran the numbers and all the rolls look quite good. All non-zero roll values occur when the plane turns, clockwise turns yield negative rolls, CCW turns yield positive turns, and the values of the rolls are reasonable. I'm marking this as solved. Thanks!


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landing with east to west wind on runway 11/29
i was wondering if anyone could help me out with this doubt i have. the kananga airport in dr congo has a single runway labelled 11/29. upon checking on google maps the number 11 is on the left end and the number 29 is on the right. if the wind is blowing east-west which of the two sides would i say i was landing on? i know planes land against the wind so i figured one would land from left to right , meaning one would land on runway 11. but after i read about runway numbering i got more confused because i understood that the number 11 should actually be on the right side rather than the left. can anyone help me understand why the runway  has an 11 on the left (which i understood should be west) and 29 on the right (supposedly east)? and if im correct to assume that with an east to west wind one should land from west to east meaning one would land on runway 11?? thanks a lot

The number for an eastward landing (11) is on the west end, since when you're landing towards the east, the west end of the runway is the first end you see.

if im correct to assume that with an east to west wind one should land from west to east meaning one would land on runway 11??

That is correct. The convention is that the runway number is the magnetic heading of the runway to the nearest 10 degrees, but with the low-order 0 removed. Thus, to the nearest 10 degrees, the magnetic heading of runway 11 is 110 degrees. That's the approximate heading the aircraft is going to have when they land on that runway. That would be a landing (or takeoff) to the east rather than the west. In other words, it's the heading of an aircraft using the runway as opposed to where the runway end is in a left-right sense when looking north.
Kolwezi Airport has only a 3° west magnetic variation. Thus, in your area magnetic heading and true heading are nearly the same.

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How long is the training to get a Boeing 777 rating if one has a 737-200 rating?
I was wondering if anyone knows how long the training for a 777 takes if one already holds a 737-200 rating.
I've looked it up online and didn't find an adequate answer.
According to this site, the course for 777 PIC initial certification takes 21 days (48 h) and the course for 777 PIC recurrent certification takes 4 days (16 h), but there's no mention of anything regarding previous ratings.
Here, someone says they think that to go from the 737NG to the 777 would only take 3/4 sim sessions, and someone else says that the course to go from 737NG to 777 is shorter than the course to go from 737-200 to 777.
Does anyone know how many days the "conversion" course would take?
Am I correct to assume one doesn't have to take the full 777 initial certification course if one holds a current 737-200 rating?
And what if the 737-200 rating is no longer valid... does one have to take the full 777 initial certification course?
Thanks a lot

A flight school can offer an "abbreviated type rating course" which takes into account skills from previous type ratings, assuming you have flown an adequate number of flight hours on those ratings.
Boeing also offers a shortened type rating course, details are halfway down their page. To answer your question, Boeing claims 13-15 days for their training and requires 1,000 hours of experience, but it does not depend on a specific rating. There is an even shorter 777-787 conversion as well, that requires a 777 rating.
I think their assumption is that they can skip a lot of basic handling, CRM, checklist type training and focus on the technical aspects. A training provider with the proper approvals could offer a similar class.
Also, for reference, there is the Airbus A320-A330-A340-A380 conversion training which uses the proprietary term "cross crew qualification", and takes less than 2 weeks versus 6 for a full rating. This is done under a EASA approved syllabus, which allows you to do it at any qualified school. Here the requirements are 150 FH on the appropriate Airbus type, and 3 months of experience.

That across the Airbus family with a high degree of commonality. The only commonality between B737-200 and B777 is the name on the bonnet and the number of wings.

@Koyovis The steam gauge 737 isn't even common to the NG: Boeing says that's a 2 week class. 737NG to 777 is much more common. Similarly, the A300/A310 is in its own world, which even had 3 incompatible variants: the 3 crew analog, the FFCC 2 crew analog, and the EFIS model.

Indeed, exactly my point. It would seem to me that the complete initial B777 type rating course would need to be done, since there is almost zero commonality.

@Koyovis I see your point. Boeing's advertising doesn't specifically call out EFIS or a type particular rating, though I do think they may be making certain optimistic assumptions. Still, basic handling and crew coordination  skills still carry over.

Thank you for the links. That  Boeing link was actually really good (though I didn't find that specific conversion mentioned -- is it under a different name?). As I don't know what the two types of aircraft have in common  (or what they have in common with other aircraft), I cannot assess whether one would need a shortened course or a full one, but    knowing that the shortened course takes 2 weeks (and the full one takes 3) is helpful because now I know more or less how long it'd take. Thank you all for your help.

is the 1000 hours of experience related to the Boeing family? Or could someone with 1000 hours on a Cessna, complete the 777 course in 15 days?

The intital or recurrent type ratings for a jet are a predetermined duration set by the school, regardless of any previous type ratings a student holds.  However the student will probably digest the material much quicker if the possess a type rating in an aircraft similar in size, features, systems, and performance, or I’d the new aircraft come from the same manufacturer, to the one they are currently seeking a type rating for.

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What are the books that a must read when starting instrument rating?
Also, is there a book like the visual maneuvers handbook regarding instrument procedures?

Instrument Rating - Materials to start reading?. Such forum is more appropriate than this Q/A site where resource recommendation is off topic.

Define “must read”. As far as I know, it is possible to get an instrument rating without reading any books.

I'm going to challenge that @Jim. Possible, theoretically  perhaps maybe, but irl nope

@Jpe61   Well, I can tell you that in the 80’s - early 90’s all I had to do was fly with my instructor and watch my Sporty’s Instrument Rating VHS tapes.  :-)

@Jpe61 - Actually I think it was John and Martha King’s tapes; Sporty’s was the new competition.

I bet you read some book though @Jim, not necessarily on instrument flying, but something like Huckleberry Finn, maybe

The best book on aviation weather for the commercial/IFR/ATP pilot I've ever read is the textbook published for Royal Canadian Air Force aircrew called the RCAF Weather Manual. Simple, clear, easy to understand aviation weather theory. Beautifully written. Can't recommend it enough.

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What are the effects of camber location in an airfoil?
Am I correct to assume that for a low speed aircraft, the location of the camber in the MAC's airfoil will correspond with the aerodynamic center? Also, does that affect the moment coefficient? And what about a high speed aircraft (M>1)?

[Does] the location of the camber in the MAC's airfoil correspond with the aerodynamic center?

No.
The aerodynamic center is where the additional lift force from a change in angle of attack can be summed up. Since camber only determines the lift portion that is independent of angle of attack, changing the camber will not influence the aerodynamic center.
For a high aspect ratio wing in subsonic flow, the aerodynamic center is at the quarter chord position. With decreasing aspect ratio this point moves slowly forward until it arrives at the forward tip for a slender body.
Also, strictly speaking, there is no "location of camber" — camber is distributed over the whole chord. What you probably mean is the highest point of the camber line, but that point has little aerodynamic significance. The local curvature of that line is much more important.

Also, does that affect the moment coefficient?

Yes.
Generally, the more an airfoil is cambered towards its rear end, the higher its pitch moment coefficient will be. This is called "rear loading" and is used in order to combine decent thickness with good lift in transsonic airfoils. You can use the lengthwise location of the highest point of the camber line as a guide how high the moment coefficient will be.

And what about a high speed aircraft (M>1)?

That depends on wing sweep. Normally, aircraft designers try to keep the wing's leading edge subsonic over the full speed range. But I guess you wonder how the camber will influence a wing's performance once it moves at full supersonic speed. Well, the bad news is that camber will drive up wave drag without any influence on lift. Supersonic aircraft, therefore, use thin airfoils with very little camber, and the camber they have is only there because it helps in subsonic flight.

No. Generally, for low speed aircrafts, the AC is located around 25%MAC(non-laminar airfoils) or 50%MAC(laminar airfoils) which, in both cases, represents the maximum thickness location not the camber(which is the maximum distance between the mean line and the chord line).
AC is the place where de Cm is nearly constant, so you can say the Cm 'doesn't change'. The momemt, obvioulsly will change with the aircraft's attitude, but the coefficient doesn't.
For supersonic flight it's, generally, AC is around 50%MAC.


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What are the effects of rotating airfoil sections to match wing's dihedral?
While designing a wing, what are the advantages and disadvantages of rotating the airfoil sections to match a linear dihedral (airfoil sections always perpendicular to the dihedral line)? Does the lift force component tilts inwards and because of that, horizontal component  cancels out and there is a decrease in lift?

All you do by not tilting the airfoil (around its longitudinal axis) is to reduce the relative thickness of the wing. Since that effect is proportional to the cosine of the dihedral angle, the effect is negligible for realistic values of dihedral.
Lift will most certainly not be affected, because it, being at its root a pressure, always acts perpendicular to the local surface.


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When calculating the lift coefficient,should I use only the lift provided by wings, or also the net upward forces of the entire aircraft?
I'm preparing a wall force analysis from an aircraft at different angles of attack that resulted from a cfd simulation, so I can plot the lift & drag polars. For the lift coefficient, should I only pick the wing surfaces to obtain the lift and drag forces or all the airplane's surfaces for the net forces? The fuselage and the wing fairings have positive lift, so I'm a bit confused if I need take them into account.

It highly depends on the size of the fuselage relative to the wings. Like if you are working a missile then there is a noticeable chance that the majority if not all of the lift is generated by the fuselage.

Welcome to Av.SE!

Why not make multiple analyses? I mean, if you already have the data, presumably you have a working algorithm you can plug in to say matlab, you might as well.
One for the fuselage, one for the wing fairings, one for the wings, one for the entire aircraft. Should give a nice spread of data to go over, figure out which parts are causing the most drag and re-design them to produce less drag, etc. Ideally you would write software to do all this for you and test a range of shapes/sizes for different parts like the fuselage/fairings, nose cone, etc.

The lift coefficient tables from wind tunnels are mathematically adjusted for  tunnel peculiarities and based on above average surface smoothness, a 2-3ft chord, and an infinite wingspan. So, they are more efficient than the real world. However, the use of these numbers ends up being a very close approximation to the real world because the fuselage and power plant add additional lift. So, use the CL for the wind tunnel and you will have a very close real world approximation. I wrote a MS Small basic program for stall speed estimates using published CL. The result was within 2% of real planes.

Well, it depends on what you want the lift coefficient for, the entire aircraft,. or just the wings? It sounds like you want it for the entire aircraft. As to whether it's worth the additional effort, that would depend on how much lift is being generated by the rest of the aircraft surfaces. Again, form your question, it sounds like, at best, you are unsure about this, so the only way to find out is to include the aircraft surfaces in the calculation, right?

Thank you everyone for your comments. The reason I'm doing this is because I need to plot the longitudinal, lateral-directional (static) stability graphs of the aircraft.

If you want the lift for the whole airplane, of course you add the contributions for each part. While the fuselage might indeed contribute a positive lift, its presence will lower the lift of the wing. In the end, all parts of the aircraft will stick together when it flies and need to create just enough lift to keep its mass airborne, so it makes sense to look at the sum of all parts.
Since the wing will dominate lift, it is common for first-order calculations to consider the wing only. However, if you have data for all parts, you should include all their contributions, too, for a more precise result.

$S_{Ref}$ could be any arbitrary area you choose .Only requirement is that it needs to be consistent for all calculations.  Most commonly used $S_{Ref}$ is the wing area.


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Could wing tilting be a mechanism for VTOL and high angle of attack for jetliners?
My question is about turbofan airliners, could an airplane tilt its wings 90 degrees so that the fuselage horizontal axis is perpendicular to the longitudinal axis of the turbo fan. This way an airliner could be VTOL and at the same time it could have even 90 degrees angle of attack. Am I missing something?

I am not much understanding, do you want it to tilt its angle at take and landing, otherwise the wings are normal?

I could have sworn we've had this question before, but I can't find it atm.

Before asking if its possible, ask if its even desirable?  What possible use could there be for a VTOL airliner?  They're nice for certain military applications, but just pointless for civilian passengers. You consider the additional expense, fuel, weight, and complexity, and realize its a solution searching for a problem.

@abelenky The user is asking if it's possible. You might not find that an interesting question, or even one worth asking, but it's what the user would like to know. The user doesn't need to read your obvious disdain for his or her naïvety.

The answer to any "is it possible" question is virtually always, "yes, with enough time, money and resources, virtually anything is possible".  Possible is not a reasonable criteria.

@AEhere I believe you're thinking of Could a helicopter fly like a plane?

What you describe is called a tiltwing. It has wing-mounted engines which swivel with the wing so their thrust points up for the VTOL mode. Building a tiltwing airliner will face two challenges:

Normally, the static thrust force is only one third of the weight force, so the tiltwing airliner needs to add three times as many engines - at least. Better make this four times as many so there is some excess thrust for hot and high conditions.
In order for the wing to tilt freely, it needs to be straight. A swept wing will hit the ground during tilting unless an absurdly high landing gear is installed. Now the thrust acts along one line - how do you control pitch? You will need another engine at the tail (or nose) which can be quickly throttled in order to balance the aircraft in pitch.

Needless to say that this cuts into payload and makes this design uncompetitive unless VTOL operation is a must.

Could? Yes.
Would it make sense and be viable? No, it massively complicates the structure, fuel system, and hydraulics. It also requires a different FCS for the transition and hover flight, as well as very heavy actuators to actually tilt anything.
The closest thing that exists nowadays are tilt-rotors like the V-22, which only tilt the engines. Other proposals in the same area go further and tilt only the rotors themselves, precisely to avoid turning a huge part of the airframe.

You also need engines far more powerful than if you used a runway, and add a stability system to control the airplane at low speed. More weight and complexity.

I addition to what Peter Kämpf has written, regarding point 1. If I unterstand the things correctly a plane needs generally a thrust-to-weight-ratio above 1.0 to be able to fly vertical. Planes are generally below 1.0 because their wings generate enough lift during flight and more powerful engines would be ineffective. For example the 747 is around 0.27 and as Peter Kämpf noted, VTOL capability would require 12 additional engines.
https://en.wikipedia.org/wiki/Thrust-to-weight_ratio
Some fighter jets have more than 1.0 which are, for example, the Typhoon or the F-15. They shall be superior over others and so efficiency is defined different. And of course also the Harrier, not designed as superior fighter, but it must operate as VTOL aircraft. While not listed, I think the V-22 also must be above 1.0 ratio. Interestingly, the F-35 is with full fuell tanks far below 1.0 but it is a STOVL aircraft, not VTOL.


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where can i get a aviation map of korea with the moes, and highways and vectors
ive been searching for a aviation map of South Korea and cant find anyone that sells it. does anyone know where I can find one for flight planning

What is "moes"?

My bad i meant to say moa

You can use edit button to edit any of your post.

Data location is off topic as defined by the help center

Those are available in section ENR 6 of the South Korean AIP. Direct link.

But you have to be able to read Korean, and Google Translate won't work because of the way they display their text. Also, the direct link is not responding.

Ah, the direct link is responding now, but about as slow as molasses on a winter day in Pluto. :)

@JuanJimenez You don't need to read Korean, just click the big button that says "AIP" in the top left corner. Their AIP follows the international standard format used in pretty much every country all over the world, so it's extremely easy to navigate.

You can get one for free here but I have no idea if it is legal to use for air navigation.


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Eviation Alice Powerplant Placement
I saw an article from the BBC about the Eviation Alice this morning that left me puzzled. The design is a tail wheel tri-motor with the power plants located in a pusher configuration at the wing tips and tail.

The article describes the aircraft thusly:

"Alice is an unconventional-looking craft: powered by three
  rear-facing pusher-propellers, one in the tail and two
  counter-rotating props at the wingtips to counter the effects of
  drag."

I'm wondering about what drives the configuration of the Alice? What compelling benefits does the configuration offer to balance the challenges it poses:

Controllability in the case of the failure of one of the wing tip power plants - even with the ability to fly (and presumably continue a worst case take off) on only the rear motor - why wouldn't you move the wing mounted engines inboard and have even better "engine" out performance?
Ground clearance and crosswind problems for the tail gear.
Designing a light and efficient wing with tip mounted power plants.

Obviously Eviation thinks that their configuration makes sense, has anybody heard or seen their arguments?

That design looks to be more aesthetic than practical. A tailwheel configuration is much more limited in crosswind operations.

I doubt this aircraft will ever fly, or be certified. There is no way to control the yaw of 350 horsepower at the end of a 25’ wing if the other engine should quit.  Even with a nose wheel,  crosswind landings will be a challenge since the wing can not be lowered into the wind.

@MikeSowsun There are other options in crosswind landings. Besides, this is not new. NASA has a similar configuration aircraft, the X-57.

@GdD -- "That design looks to be more aesthetic than practical. A tailwheel configuration is much more limited in crosswind operations." -- a tailwheel configuration is no problem as long as you can do a nice wing-low slip to cancel drift -- oops, never mind!

I suppose the intended benefits have something to do with the propwash cancelling the wingtip vortices.  Maybe they could have better gone with a tricycle gear configuration with an extra rear gear to protect against prop strike.

@quietflyer Just keep lengthening the gear until the props stop hitting :-)

According to a recent article from the travel blog The Points Guy (whose reporters were at the Paris Air Show and toured the aircraft,)

[W]hile it has three motors, the aircraft can continue to fly with just the rear motor engaged. If one wing-mounted motor fails, the other will be disabled to maintain stability, and additional electric power can be diverted to the rear motor so the plane can continue its flight.

Multiengine aircraft date back to 1913, when Igor Sikorsky's Russky Vityaz first flew. Darn thing was huge, had a balcony, of all things, and could only do 55 kts, but it was impressive.
All multiengine pilots have to pass training on what to do if an engine fails, including what happens if airspeed falls below Vmc (minimum controllable airspeed) and how to keep the airplane flying reasonably straight until you can land. This aircraft is no different. It's also possible this aircraft can at least maintain straight and level flight on just the rear engine after shutting down the function wing engine if needed.

Perhaps I should revise my question - even if the design is certifiable, it seems like they’re taking on unnecessary control problems. I’m wondering what the compelling benefits might be.


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Definition of Critical Part in type certification process
first time here :)
I'm part of the team at Flytrex (flytrex.com) and we're working with the Aircraft Certification Office to certify one of the first UAS systems (drones) under new initiative to certify drones for commercial use.
We are currently at the stage that we need to define our system critical parts.
From what I understand, a Critical Part is defined as a single point of failure that will lead to a crash of the aircraft.
Since every component in our system includes redundancies, is it reasonable to claim that we don't have any Critical Part?
Some examples:

Single motor loss - the aircraft uses 6 motors and can sustain a single motor loss.
Multiple motor loss - an independent parachute system is deployed, taking the aircraft to the ground safely.
Flight controller loss - again, parachute will deploy.
Parachute system failure - the flight controller will bring the aircraft back to safe landing (abort mission).

I think we need a clearer understanding what is considered a critical part, and then try to define the right parts in our system.
Many thanks!

Hello Amit Regev, interesting question! Welcome to Aviation.stackexchange.com

Electrical failure?

@JuanJimenez the parachute system is independent and monitors power outage, so the parachute will deploy in such case.

How does it monitor power outage? Is there a mechanical failsafe? What if the failsafe fails? Assume Murphy is watching and will throw everything at you. :)

You should check in with the certification office regarding the critical part definition, as the wording is important. In general, I would use a definition that designates every part whose failure in service could (not will) cause a catastrophic failure of the aircraft. This includes multiple failures and the degradation of the margins of safety of other critical parts.

Also, what @JuanJimenez said. How reliable is your power monitoring system? What are the effects of a false positive? What about your structure? Surely that is critical by any definition.


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Is it possible to obtain an STC for single-pilot ops in a jet certified for two pilots?
I have been doing research on whether or not one could realistically obtain an STC for a super-midsize jet certified for two pilots and legally fly it single-pilot.
For reference, my chosen aircraft is the Hawker 4000, which can be had for significantly less than most jets certified from the manufacturer for single-pilot ops. Additionally, the Hawker boasts performance figures that laugh at all other light jet numbers and its Relative Value Analysis (RVA) is comparable to jets costing $25M+.
I cannot find information anywhere on gaining single pilot certification in any aircraft not originally certified for single pilot operations. I would even take into consideration the possibility of a complete retrofit of the flight deck with Garmin 3000 or some other system designed to be flown single pilot (like that in the Phenom 300).
Thanks!

Even if you could get it STC'd for a single pilot, nobody will insure it. The other problem you have is if the Garmin 3000 is STC'd for the Hawker 4000, which I don't believe that it is. The Hawker 400 is STC'd for the Garmin 5000, but not the 3000.

Anything is possible, but feasible is another story. As has been pointed out, even if you manage to spend enough money (likely orders of magnitude more than the cost of the bird itself) to convince the FAA to do that, it will likely be impossible to insure. This is really a question asking for opinions, so I vote to close for that reason.

It's possible that cockpit width is an issue.

Cessna did this sort of thing for the CE-525 series (CJ) jets. Effectively the required crew varies based on cockpit arrangement/equipment. As per the type cert

Minimum Crew for all Flights (see NOTE 5 for cockpit equipment/arrangement restrictions):
One pilot (in the left  for all Flights pilot seat) plus additional
  equipment as specified in the Kinds of Operations Equipment List
  (KOEL) contained in the Limitations Section of the FAA Approved
  Airplane Flight Manual (AFM)
OR
One pilot and one copilot

Note 5 for reference:

Approval for operation with a minimum crew of one pilot is based upon
  the cockpit equipment installation and arrangement evaluated during
  FAA certification testing. No significant changes may be made to the
  installed cockpit equipment or arrangement (EFIS, autopilot, avionics,
  etc.), except as permitted by the approved MMEL, without prior
  approval from the responsible Aircraft Certification Office.

However that is the original type cert and it was done by the manufacturer, who has the resources to test these kinds of things. Doing it on your own is bound to be cost prohibitive, even considering you may have to pay a second pilot to fly with you for the entire ownership of the aircraft.
This is also fairly similar to the case in the heavy metal world with the dropping of flight engineers and navigators when control was consolidated to the two pilots. So far as I know none of this was done via STC but appeared on the TC itself.
Technically it may be possible, practically, it seems unlikely.
If your looking for single pilot jet performance there are plenty of early era ex-military jets hitting the market that are far faster than any business jet you can buy...


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When a hold is distance based, do you still used a timed entry?
Let's say you have to do a parallel or tear-drop entry to enter this distance-based holding pattern.  Do you still fly outbound from the fix for one minute and then make your turn inbound to re-intercept the fix or are you taking it to 4 NM?

Welcome to Av.SE

This is ultimately up to pilot discretion.

I haven't found any specific guidance for hand flying entries to RNAV (RNP) holding patterns. But part of the concept of RNAV is that the navigation system can provide the complete desired path including hold entries.
Referring to RTCA DO-236C RNP MASPS and the requirements for holding operations, I find the following: The dimensions of the hold pattern are defined based on altitude, airspeed, bank angle, and an allowance for wind. The length of the inbound/outbound leg is defined by the database or entry. In your case this is 4 NM, and is defined as d1. The width of the pattern (d2) is computed based on the above variables.
Without the details, we can still make an approximation. If you're flying at 120 kts and performing a 2 min turn (1 min for 180 deg), your pattern width (d2) would be about 1.3 NM.
The rules include the following statement:

When performing an entry into the holding pattern, the navigation
  system is permitted to use an inbound leg length (d1) that is equal to
  the holding pattern width (d2) instead of the length defined in the
  database or from pilot entry.

So the end result is that your entry cannot exceed the 4 NM leg length, and could have an inbound leg length as short as 1.3 NM.

It is possible to reach the holding fix and enter the hold using any of the three entries described in the AIM, fly outbound on a specific heading for a given length of time, and then turn inbound and intercept the inbound course exactly at the point which is the required along track distance from the fix, or required inbound time to the fix.

The question as written is regarding a hold with distance. As you have stated, the FAA and ATC has recommendations on how to enter the hold. Ultimately, they leave the entry to pilot discretion as long as you stay within the designated protected area. The leg length being an area limiting factor. If you choose to cut your entry leg short after initial passing of your fix by reducing the leg length, you are allowed to do that. If you choose to use timing as a basis to make your turn back to the fix, you can do that also. A 2 minute leg prior to a standard rate turn back is a good practice.

In either the distance or timed scenario, your turn back must begin prior to or up to reaching a 4 nautical mile DME (or RNAV) distance from the fix of the IAP depicted in the question. Some profile views will also have a distance restriction for the hold and/or procedure turn to which you must adhere.

No the hold makes use of a straight line distance from a GPS identified waypoint and is not timed.  This can also be done with holds on a VOR/DME and or VORTAC using a DME or GPS distance from the Navaid.

Do you have a source for this? (Also note: the question was about the entry, not the actual hold itself)

See the FAA’s Instrument Procedures Handbook.  If you have the equipment that can measure distances (which would be required here in order to fly that approach) and the hold is published with and outbound leg defined in a straight line distance from the hold point, there’s no point in timing the outbound leg.

Yes there is a point: expedite the approach. A hold entry in lieu of procedure turn would be expedited at lower speeds—90 kits for example—by flying a one minute teardrop entry leg rather than a 4 NM leg. The time saved would be approximately 5 minutes in this case.


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Why does the effective alpha/CL increase at wingtip of swept wing?
If anybody could explain this in an intuitive way I'd really appreciate it. I haven't been able to find a clear answer on this. I understand that the increase in effective AoA/CL and the spanwise flow & resultant thickened boundary layer at the tip is responsible for the tip stall (if uncorrected for in the wing design), but I can't wrap my head around the reason why the effective AoA and CL increase in the first place. On a rectangular wing they decrease because of the upwash induced from the vorticies, so why doesn't it have the same effect on a swept wing? Most of the explanations I've found on here are geared for aerodynamicists and not the average joe pilot like me, and the explanations given in most pilot resources are overly simple if an explanation at all.
The book "Handling the Big Jets" by D.P Davies says "Too much sweep produces poor oscillatory stability and a tendency for the tip to stall, causing pitch up" and that's it..
ACE the Technical Pilot Interview by Gary V. Bistrow says "A simple swept and/or tapered wing will stall at the tip first [...] This is so because the outer wing section produces a higher wing loading due to the wing taper, which causes a greater angle of incidence to be experienced to a degree where the airflow stalls at the wingtips. The boundary layer layer spanwise airflow, also a result of sweep, further contributes to the airflow stalling at the wing tips."
This ATPL CBT video I watched says the airflow separation starts at the tips because the weak wingtip vortices allow separation earlier than the inboard part of the wing where the stronger vortices from the wider chord reduce the effective angle of attack... what?!
I have an interview coming up where this question could be asked, and I don't feel like I'd be able to deliver an answer with any sort of confidence at this point.

You are reading the wrong sources.
I've never come across the claim that the lift curve slope of a swept wing increases towards the tip. That might happen in ground effect (when the tips are very close to the ground from the swept wing's incidence), but that effect is very weak, or when the wing tip has noticeably less sweep (crescent wing).
But not on "regular" swept wings in normal operation.
If, however, the intended meaning of your question is to ask why the tips of backward swept wings reach stall conditions earlier than the inner wing: That they do indeed. Read on for an explanation.
What your sources describe is foremost caused by the thickening of the outer boundary layer of backward swept wings. This is compounded by aggressive taper - but this taper effect is very similar in unswept wings, too. Next, when you plot aspect ratio over sweep angle, you will find a boundary above which swept wings develop undesirable stall behavior, so the wing's aspect ratio is another factor (which your sources stay mum about).
To prepare for the interview, I advise you to read some of the answers here, like:

Swept wing layout for maximum lift at transsonic Mach numbers.
Sub- and transsonic flow separation.
Stability of swept wings.
Why the better designs use increased sweep at the tip, precisely to lower the local lift curve slope.

And make sure you follow the links in the linked answers as well if something is unclear.

Thanks Peter. So in simplistic terms, as AoA is increased, the spanswise flow starts to pile up towards the tips which increases the thickness of the boundary layer, making it more susceptible to separation. So when the AoA of the plane is increased further, the separation begins at the tips first because of its thicker BL. But it does not in fact have a higher effective AoA or CL since the airflow over the top of the wingtips doesn't have a larger pressure differential than the root (which makes sense - it has less lift anyways..) hence causing the tip stall. Is my understanding correct?

@Speedalive: Yes, your understanding is correct.

Peter, I was initially confused as well, but it seems @Speedalive didin't mean the lift curve slope. He just wrote 'AoA/CL' because both are involved (and in the linear range can be used interchangeably). The standard idea for simplistic (just so as not to say bogus) explanations like quoted is that the [effective] AoA is somehow higher at the tips, but it doesn't imply that the slope is different. In my experience, pilots (and their sources) rarely understand the lift curve slope as a specific parameter at all. They know things are normally linear, but that's about it.

@Zeus: Thank you, that makes sense. So the interpretation of "the effective alpha and/or c$_L$ increases towards the tips" should read "the tips get nearer to stall". That they do indeed.

Thanks for the clarification!

I just found this answer after seeing this Reddit post, and it completely answered my question! Wish I could give +2.


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How does the A350 variable camber system decrease drag?
Just curious how the A350 reduces drag/fuel burn by extending the flaps ever so slightly during cruise. Everything I've read about extending flaps says that extending them pushes the centre of pressure rearwards. This causes a pitching down moment, which means you need to increase the AoA, but this affects drag and fuel burn negatively (a pitch up command increases the down force from the tailplane which has the result of increasing the effective weight of the plane, which requires an even higher AoA to compensate, which also increases the induced drag).
Here's how it works as per the (A350 Flight Deck and Systems Briefing for Pilots)

Differential Flap Setting and Variable Camber
The Differential Flap Setting and Variable Camber enable to optimize the loads and drag on the wings.
  Small flaps deflections (4° maximum) either symmetrically or asymmetrically, enable to automatically:

Optimize the wing camber to reduce wing loads and drag
Perform an optimized Lateral Trim function.

I don't remember the details exactly, so I'll post it as a comment. However, I think the reasoning makes sense. I once heard a talk about this while I was an intern at Airbus about this topic, and I if remember correctly it was mentioned that instead of rotating the entire plane to reach the correct angle of attack of the plane during cruise, they would extend the flaps slightly. By doing this, it was possible to keep the fuselage more horizontal, and reduce the drag resulting from the fuselage.

Just complementing the answer from @ares which is quite good and refers to the main effect. I would like to refer to another "secondary" effect that implies also drag reduction.
When designing an airplane the structure is designed taking into account several factors, one of them, is the maximum load that the airplane can be exposed to.
Airbus has designed a system that during flight optimises the loads over the wing. Let's say, for example, that, without the variable cambering system we have a determined maximum load (let's say A). Using the variable cambering system the airplane is capable of reducing the load A (maybe increasing drag) to B (being B < A).
So, when designing the structure, assuming that the variable cambering system will be used, the airplane will use B as design point and not A. As B < A the size and weight of the structure with the variable cambering system will be lighter.
A lighter structure will imply less lift force needed and so less drag produced to achieve such lift. So, from a design optimization point of view, the variable cambering system is, essentially, providing new design variables which will allow a better opitmization of the airplane, reducing the drag.
Essentially, @ares has properly described the "active" mechanism, and I have described a "passive" one.

Thanks Trebia! Still not sure if I'm getting the basics. If you could have a look at the questions I asked Ares I'd appreciate it!

Your 4th paragraph looks like it got cut off: "As B"...

@FreeMan , thanks, I used a simbol actually used everything behind as a comment.

To answer your specific question: The specific ability of the A350 is to extend the inboard flaps independently of the outboard flaps and to extend them by only a few degrees.
The A350 has a high wing sweep of 35° and a high wing span. The result is, that seen from the direction of flight, the outer parts of the wings lie far behind the inboard flaps. This means, that the center of lift of the wings is actually located around where the inboard flaps are. If the inboard flaps are extended by up to 2° (the maximum possible during cruise flight), the additional lift created by them will be located close to where the center of lift was already before!
I think, this can be seen rather easy from figure 2 of the design paper linked by Speedalive. The blue line and label "inboard flap" shows where those flaps are, that can be slightly extended. Keep in mind, that those flaps are actually not just at the edge of the wing, but start already a meter or so before. If the flap is rotated by up to 2°, it produces additional lift along all of its surface, not only the edge. So the center of additional lift created by the inboard flaps alone might even be slightly before the center of lift of the whole wing.
Furthermore, as already said by others, this slight extension of the flaps by 2° creates more uplift and thus allows to reduce the pitch angle of the whole aircraft by maybe 1°. This reduces the angle of attack on the wings by the same amount, and with the typical asymmetric wings used on aircraft today, lowering the angle of attack usually leads to a slight forward movement of the center of lift.
So, putting it all together: Additional uplift is created around the back of the inner part of the wings. In exchange, the angle of attack is reduced and less uplift is created on the outer part of the wings, which actually lie behind the (slightly) extended flaps, and also on the front parts of the inner wings. Both reductions have a combined center of lift reduction, which is close to the center of the additional uplift from the flap extension.
What others have already pointed out: At the beginning of the cruise flight, the fuselage is usually heavy, as the center tank is still filled with fuel. So more force needs to be transferred from the wings to the fuselage. If the additional force is created by increasing the angle of attack (or by flying at a lower altitude in denser air), it is created along the whole length of the wings and the whole structure needs to be able to carry that additional force over the full length of the wings. With the slight extension of the inboard flaps only, the additional uplift is created on the inner wings instead, close to the fuselage. So the maximum weight, that the outer wings have to carry for extended periods of time, is reduced. They don't need to be so strong, so less material can be used there, saving on aircraft weight.
Trim Tanks
I tend to disagree with others, that claim, that this "cambered wing" is a good replacement for the trim fuel system in A330 and A380. They are different systems to optimize fuel efficiency: The "cambered wing" is about reducing the angle of attack during heavy flight phases. It actually can also be used later, for example, when ATC asks the pilot to climb 1000 feet higher than what would be optimal for the current weight to avoid other traffic. Then again, flying higher, the aircraft would either had to increase the angle of attack and thus also the drag, or use the slight extension of the innermost flaps as a more efficient alternative!
Trim fuel, on the opposite, is about managing the center of gravity and let it get closer to the center of lift. When flying close to ground during initial climb and final approach, a large separation between these two points is essential for stable flight, especially during upsets like crosswinds and similar. During cruise flight, less separation is required and trim tanks can help to reduce it. During take off and landing, even more separation is required, and this is achieved by extending all flaps (not only the innermost ones) by a large angle (and not only by 2°), which substantially moves the center of uplift backwards.
On the other hand, the high wing sweep of the A350 allows also some trimming by exploiting the different fuel tanks of the wing only: Pumping fuel to the wings' outward tanks (or just consuming the fuel from the inner tanks, first) will move the center of gravity substantially to the back.

Briefly speaking, there at two types of drag: profile drag and induced drag. Profile drag can be said to be composed of viscous drag and shock-wave drag (these two often interact). Induced drag is a result of the local effective velocities on the wing due to the -vortex- circulation. I don't know what is your background and thus I won't go more in depth. I will just mention that, if you want to simplify things, profile drag can be seen as the drag that a 2D airfoil would have, while induced drag is a measure of the efficiency of a wing geometry. In reality, these two types of drag may strongly interact when the aspect ratio of wings is small.
For example, it is well known that elliptical spanload distributions are optimal (when induced drag is the objective). This 'elliptical spanload' can be reproduced if you design a wing of i) constant chord with elliptical distribution of twist, ii) elliptical distribution of chord with constant twist, iii) a combination of twist and chord that gives an elliptical distribution.
What AIRBUS says, is that we can vary the chord in-flight to optimize spanload and root-bending moment. What they actually do, in simple terms, is to optimize for induced drag and possibly control, i.e. stability and flight controls by changing the chord distribution. Now, if you change camber, you will also optimize for profile drag (because camber is a property of the 2D airfoil) and twist (in the sense of aerodynamic twist). For commercial jets, flying in transonic conditions, this means that the camber is modified to reduce shock-wave drag loss.
Clarification: Aerodynamic twist refers to changes in camber along the span. It is often more efficient to change the airfoil shape along the span instead of using the same airfoil at a different angle.
If you want more details, please let me know.

Thanks Ares. I'm still not sure if I completely understand. How is the A350 is able to reduce induced/parasite drag by extending flaps/increasing camber? My understanding is that flaps increase downwash which decreases the effective alpha of the wing, giving an additional, downstream-facing, component to the aerodynamic force acting over the entire wing which tilts the lift vector back and causes induced drag. Same with the slight increase in wing chord/area + decrease in aspect ratio.. As far as I've read in pilot resources, extended flaps don't add to fuel efficiency at all.

In the A330 they have a stab trim tank which keeps the CG within 2% of the aft limit and I can see how this saves fuel.. It is well known that an aft CG saves fuel due to the reduced tail down force (effectively making the plane lighter) which reduces the required alpha for a given weight. This reduces induced drag, increases TAS, and decreases fuel burn. With the A350, I understand that they burn the center tank fuel first for a similar effect, but the slight extension of flaps moves the CP rearward and increases downwash which, from what I understand, has an opposite and negative effect.

Obviously, this is not the case in reality, because if it had a negative effect, they wouldn't have designed such a system. So I'm curious what concept I am misunderstanding or missing. Currently studying for my ATPL exams and I keep getting hung up on this question when I'm reviewing principles of flight. I won't be asked this on an exam or anything, but it bothers me that what I'm taught in ground school can't explain how this system improves the 350's aerodynamics.

@Speedalive Too many comments/questions! I'll try to answer them... So, "How is the A350 is able to reduce induced/parasite drag by extending flaps/increasing camber?" I think I have described the whole process in my answer. If you are not convinced I have to bring the equations in, but I guess this is too much. The thing is that this induced velocity field depends on what's happening on the whole wing. Thus, playing with the camber and chord distribution you can 'tailor' your spanload to each specific flight condition.

@Speedalive If you wanna understand this better, a good way to start is Prandtl's lifting-line theory, assuming you have a basic background in aerodynamics.

@Speedalive I am not fully aware of the details of the plane. My answers face the problem from the general perspective of aerodynamics. For any configuration (position of tail, fuel tanks, engines, fuselage moments, blah blah blah) everything boils down to what are the optimal coefficients of lift, drag and moment.

@Speedalive For a particular configuration you constraint your wing to operate at a specific moment coefficient, which again depends on the spanload. So, in flight, as the CG may move because of the fuel burned, the required moment coefficient changes. This can be leveraged if you can modify the camber/twist of the plane in-flight to get a more efficient wing. Now, the 'optimal' is a trade-off between total drag, stability considerations, range of motion of the mechanical parts (flaps) etc.

@Speedalive I said "everything boils down to what are the optimal coefficients of lift, drag and moment", I meant that for the coefficients of the wing, as the mechanism you are asking for is an active control on the wing (stack won't let me edit the comment after 5min have passed...).


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Are rhumb lines ever used as leg paths for RNP/RNAV?
Are rhumb lines ever used by an FMS as leg paths for RNP/RNAV? If so, what arinc 424 leg type would use them?

No. The FMS does not use rhumb lines.
All RNP/RNAV flight path legs are geodesics. The only exceptions are RF legs (a constant radius circular path about a turn center) and hold legs (a closed racetrack pattern).
To quote RTCA DO-236C, Minimum Aviation System Performance Standards: Required Navigation Performance for Area Navigation:

RNP Routes and user-preferred trajectories are assumed to use a series
  of fixes. The desired path is defined by a series of geodesic tracks
  joining successive fixes.

It also includes the following definition:

GEODESIC LINE [Bowditch]
  A line of shortest distance between any two points on a mathematically
  defined surface. A geodesic line is a line of double curvature and
  usually lies between the two normal section lines which the two points
  determine. If the two terminal points are nearly in the same latitude,
  the geodesic line may cross one of the normal section lines. It should
  be noted that, except along the equator and along the meridians, the
  geodesic line is not a plane curve and cannot be sighted over
  directly.

thank you very much


text:
What are the fields of the ground reply (uplink) to a weight-and-balance request?
I am trying to send from the Aircraft MCDU AOC menu (via ACARS Data Link) , a Weight/Balance Request downlink to the Ground Station .
I would like to know what are the fields that make up the uplink message to reply to this request ?


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Why do planes need a roll motion?
Isn't only pitching up/down and yawing right/left enough for travelling from one location to another?

Possible duplicate of Are the functions of ailerons and rudder similar?

So Tex John can do this in a 707 and sell some airplanes.

Be aware, Yawing does not, by itself, turn the aircraft. All it does is cause the aircraft axis to rotate. The velocity of the aircraft does not change direction from the yaw itself. The yaw does however cause the relative wind to hit the side of the fuselage, causing a sideways force  (fuselage lift), that does cause the velocity to change direction.

"enough for traveling from one location to another"...  Yes, under ideal conditions a skilled pilot could safely complete a flight from point A to point B.  (just like one day in my 20s I drove my car home from work steering with only my knee to prove that I could...)  But that's different than "maintain adequate control under a wide range of flight conditions", to which the answer is a resounding no.

This answer is specifically for a fixed-wing airplane, not a helicopter or multicopter, although some aspects of it will apply to them too.
What is a turn?  A turn is a curvature in the flight path.  This means a turn is a form of acceleration.  Specifically, a centripetal acceleration-- the flight path is being continually "bent" toward the center of the turn, creating a circular path.
Newton's second law is force = mass * acceleration, or acceleration = force / mass.  You can't have an acceleration unless you are exerting a force to cause that acceleration.  You can't have a centripetal acceleration unless you are exerting a centripetal force-- a force vector pointing toward the center of the turn.
What part of an aircraft is designed to produce force as efficiently as possible?  The wing.  If you roll the aircraft into a bank, you tilt the wing's lift force to one side, and you've now a created a centripetal force as efficiently as possible-- i.e. with the least drag possible.
Instead of banking, you can also turn by yawing the fuselage sideways relative to the flight path to expose the side of the fuselage to the airflow.  Essentially this makes the fuselage act as a sideways wing, generating some sideways lift or "sideforce".  This method also points the thrust vector from the engine or engines sideways relative to the flight path.  So this method will create some centripetal force, but it is very inefficient-- drag will be very high.  It will also be uncomfortable for anyone in the plane as they will tend to be thrown against the "upwind" side of the fuselage.
Note the difference between causing or allowing just enough yaw rotation to let the nose track around the horizon and "keep up with" the changing the direction of the flight path as the flight path curves due to the centripetal force generated by the banked wing, and forcing the nose to yaw to point in a different direction than the aircraft is actually travelling through the air at any given instant, as would be required if you were trying to turn without banking at all.
A car on flat ground can turn without banking fairly efficiently but that's only because the tires usually grip the road quite well, so it only takes a very small sideways angle between the alignment of each wheel, and the instantaneous path of travel of that part of the car, to generate a strong sideways force.  So you don't scrub off a lot of energy due to excess drag.  Very different from dragging a fuselage sideways through the air at a large sideslip angle.
A good source for starting to learn about the basic physics of flight is John S. Denker's "See How It Flies" website .  On second thought, this website is not so much designed to introduce a complete novice to the basic principles of flight, as to help someone with some flying experience and training understand the "why" of it all on deeper lever than some pilots ever manage to reach in a whole lifetime of flying!

Comments are not for extended discussion; this conversation has been moved to chat.

Might be worth adding to your answer that the primary function of the rudder is not to steer the aircraft (as one might intuitively think) but to co-ordinate turning flight and cancel out any lateral aerodynamic forces by keeping the aircraft pointed into the relative wind. It has a secondary function to intentionally create a sideways force in a maneuver called a sideslip used to either bleed energy for landing or keep the aircraft pointed down the runway during a crosswind landing.

technically, a car does indeed bank slightly, by compressing its suspension somewhst when in a turn, but it's so little you hardly (if at all) notice.

@jwenting -- but doesn't it tip in the wrong direction for passenger comfort, i.e. toward the outside of the turn?

@quietflyer centrifugal force :) And you can't really notice it in a car anyway unless you go at ridiculous speed through a very sharp turn. Ask Richard Hammond how that goes, lol

For what it's worth, most missiles, which have four wings sticking out at 90 degree angles, do NOT need a roll motion, they just deflect the appropriate wings to generate the force vector in the direction they want to turn in.

@jwenting, a car does not “bank” while turning on a level surface, it “lists” to the outside of the turn which is the opposite effect.  (And depending on the car and driver, it may be very noticeable without requiring “ridiculous” speed!)  Freeway on ramps and high speed racetracks are banked to compensate for this effect.

In addition to the use in turning, you need to realize that the atmosphere is not static.  It's in constant motion, both horizontally and vertically.  The vertical currents can be strong enough, and localized enough, to tilt one wing or the other*, thus causing an involuntary roll.  If you don't have some means to counteract this induced roll, you now have a plane that is highly unstable and likely to crash.
*Flying a sailplane or small power plane on a day with lots of thermal activity is a constant balancing act, using small control motions to keep the wings level (or at the bank angle you want) despite what the atmosphere wants to do to you.  And let's not even get into the sort of vertical currents you can get from storms...

Forget while at altitude. During one training flight, while doing touch-and-gos, I had an upset early during climbout which resulted in one wing dipping considerably. I don't dare guess by how much, but it caused a significant bank, and we were at a few hundred feet or so AGL. Let's just say I was happy to have the ailerons then!

Think about the airplane as a body you want to control. That object has 6 degrees of freedom, including rolling. If you remove rolling you will not be able to control airplane's roll. Imagine a situation, when an small mass is over the wing and the airplane starts to roll and you are not able to compensate it.
Saying that, the way to control yaw is to use the vertical plane, and, the vertical plane will generate roll that you will need to compensate. So yaw and roll are coupled.
On the other side, the preferred way to perform a turn in the air is using roll, not using yaw. It is more efficient and faster (in space) than using yaw

Roll can be controlled indirectly by wing dihedral. The Fokker "Spin" only had elevator/rudder for control, and could successfully fly around the Haarlem cathedral in 1911.

Even if it's perfectly balanced it'd be hard to maintain a heading, and impossible to change it and remain level. There's some suggested reading... I'd suggest KSP, because roll control failure is precisely the reason I didn't get to space today in my SSTO (and because KSP is the only reason I understand any of this).

@Mazura Likewise radio-controlled sailplanes like Gentle Lady, Radian do well with no ailerons but that's because ample dihedral gives lots of slip- roll coupling for good roll control based on rudder inputs.  No problem doing steep turns, sustained inverted flight including some turning maneuvers, and even full 360-degree rolls (at least w/ Radian with motor on for extra airflow over tail).  "Super Floater" ultralight sailplane (original version) is a full-sized counterpart; later version had flatter wing and ailerons.

Here's simple way to conceive it.  An airplane turns by tilting the lift force from straight up to angled off to one side, and to do this, it has to bank.
Imagine the SpaceX rocket hovering.  If it wants to move sideways, it tilts its thrust line off vertical.  With the thrust line now pointed at, say, 11 O'clock instead of 12, some of the thrust is now creating a lateral force and the rocket moves sideways.  To keep from descending while moving sideways, the total thrust has to be increased to compensate for the portion of thrust now working to move the body sideways.  But in any case, it's the tilting of the thrust line that makes it move sideways.
The airplane's wings are making thrust (by redirecting a large package of air above and below downward) as it moves along.  If you tilt the wings, it's like the rocket tilting.  The thrust being created by the wings is tilted by the amount of the bank angle, and part of the wings' thrust or lift is now creating a lateral force, making the airplane move sideways.
But the airplane isn't hovering, it's moving forward while this is going on. Because it's moving forward, the sideways movement created by banking the wings results in it moving in a horizontal arc.  That's a turn.  Like the earlier rocket analogy, to keep from descending the airplane has to increase its total lift thrust to compensate for the thrust lost by using it to move laterally.  So you have to pitch up a bit to make more total thrust (lift) from the wings as you turn or you'll descend.
Helicopters are the same.  If it's hovering stationary, and you want it to move sideways, you tilt the rotor disc to angle the thrust line of the rotor.  If the helicopter is flying forward while this is being done, it moves sideways and forward at the same time, creating a horizontal arc, or a turn, just like the airplane.
Turning by skidding is possible also, but is also very inefficient.  Hovercraft, which are really just ground effect airplane/helicopters, turn this way because they have no choice; they simply can't bank when they need to stay so close to the surface just to stay clear of it. A hovercraft takes forever to turn because it has to skid sideways and use the lateral force of it's engine thrust to make the change in direction and this really limits their maneuverability.
However, if you drive your hovercraft on a bicycle race track with banked corners, at the right speed appropriate to the bank angle, you just have to keep driving it "straight", that is, just keeping the tail end lined up with the nose end, and the hovercraft is changing direction without having to skid.  In an airplane, you bank the lifting force instead of the supporting surface and just work on keeping the tail lined up behind the nose (ball centered), and around you go. Not a perfect analogy by any means but sufficient to be able to visualize the forces at work.
There ARE times in flying where you use skidding to make changes in direction - mainly formation flying, which includes glider towing.  When close to another airplane, it's not a good idea to move closer by banking toward the airplane you're formating on.  You maneuver by skidding, taking advantage of the inefficiency of skidding maneuvers you might say, to able to make subtle position changes will less risk of running into your formation object.  Glider pilots also learn to adjust lateral position by skids and minimal amounts of bank where needed.

A fixed wing aeroplane can also turn by deflecting the rudder. Helicopters as well. So why do they require roll for turning, would still be the question.

An airplane's wings do not generate thrust. Wings generate lift, but lift is not thrust (in typical nomenclature). Spacecraft are hugely different from aircraft in that they typically operate in an environment where there is very little to nothing to push off against, and therefore they have to rely on thrust. (Compare spaceplane designs like the space shuttle orbiter.) I usually like your answers, but conflating lift and thrust doesn't seem likely to help someone who is struggling to understand why airplanes use roll, even though there are cases where they can be oriented along the same axis.

@Koyovis: Because banking is a much more efficient & controllable way of turning.  You might ask yourself why curves on highways (well-designed ones, anyway) are banked, or why motorcycles & bicycles "bank" when turning.

@aCVn yes it's thrust.  It's action/reaction of air being displaced.  A propeller makes "thrust", from a column of air accelerated aft.  It's just a wing going round and round, inducing air to move.

Recommending maneuvering in close formation by skidding alone is not a good general recommendation.  All 3 axes are in play at all times, you just need to use small smooth corrections, but rolling is certainly an option.

If you're some distance fine, but I would never bank to get closer to someone in close formation, as in 20-50 feet away (I've done it) and a lateral movement could make you hit them. The problem is by banking you effect a heading change and put yourself on a converging path.  To stop the convergence you have to bank the other way to get back to a parallel path.  The input/result loop is way too long.  Gentle wings level skids when close.

@John K: Though I expect most of us have more sense than to get that close to another airplane :-)

John, for some reason I am just now seeing your last comment above. I have to disagree with your technique of using rudder alone to maintain position.  I was trained by the best and have a lot of experience here, (I have even flown about a 0.3 as sole manipulator in the slot position with the Blue Angels during a practice...) and lateral distance is best controlled by the stick. There really isn't any perceptible "input/result" loop delay in real close.  You just make constant, instinctive, small control movements to stay in position. Stay off the rudders, they just twist you and add drag.

@MichaelHall I learned that technique from a book, the title of which is now lost in the mists of time. Or maybe it was an article in FLYING.  Probably 40 years ago.  I take your point.  I wonder if the snappy controls of a jet make that a more desirable method than skidding an airplane with a slower roll rate.  Flying in-tow on a glider, (a form of formation flying) the effects are exaggerated by the glider's ponderous roll rate, and I use rudder only for very fine lateral corrections, with a bit of roll for larger ones, because the input/result lag from roll is significant.

Interesting perspective, and I can definitely see that technique could be airframe specific.  Thinking more, I remember besides not being encouraged, rudder use was actively discouraged.  And smaller jets are pretty responsive, little sensation of actually changing heading when fairly close, more like a feeling of just banking slightly, and sliding left and right laterally using the stick. So yeah, I think you are correct in that regard.

Pitching up/down and yawing left/right would be adequate for a high wing dihedral design, were it not for the need for cross wind control.  A high wing dihedral will make beautiful banked turns simply by pushing the rudder.  The dihedral banks the plane away from the rudder induced slip.  Great for indoors models.
Unfortunately, in a full scale outdoors (pilot in the plane) cross wind landing situation, this design would be very difficult to control without ailerons to bank the plane into the wind (in order to prevent lateral drift), while the rudder maintains ground heading.
Without ailerons, this type of plane would be easily rolled and blown sideways out of control, unless it was landed directly into the wind.

If a plane is moving e.g. north and one wants it to be moving northeast, one must apply an eastward force to the airplane.  A plane has four kinds of forces acting upon it--thrust, drag (force parallel to motion), "lift" (force perpendicular to motion), and gravity.  The direction of gravity can't be changed, and drag will be parallel to the direction the plane is moving, rather than the direction one wants it to go.  Thus, the only two forces that can be used to change the direction the plane is moving (as opposed to pointing) are thrust and lift.
If one didn't turn very fast, one could change an airplane's direction of motion while keeping the wings "level" side-to-side, but one's ability to do this in most planes would be very limited because thrust is small relative to lift, and wings are designed to produce lift from relative air motion in a fore-to-aft direction, rather than sideways.  Consequently, it wouldn't be possible to change the direction of thrust very much relative to the direction of motion without rendering the wings ineffective.  Banking makes it possible to change a plane's direction much more effectively because rather than using a tiny portion of the relatively small thrust force for that purpose, it uses a much greater portion of a much larger lift force.
As mentioned by Quiet Flier, the side profile of the fuselage may act something like a wing with regard to air which is moving past the plane at a sideways angle, and thus provide some sideways acceleration, but since most planes' bodies aren't designed to be used in this fashion, they will be far more effective at producing force perpendicular to air movement than would be wings which are actually designed for that purpose.

This answer could be expanded to mention aerodynamic sideforcec which is sideways lift generated by the fuselage acting as a sideways wing.  Even gliders with ample fuselage side area can do wings-level skidding turns, though it is horribly inefficient, and the turn rate is low, and the maneuver invites a spin if the airspeed is allowed to drop too low.  With a typical light plane I would assume that an unbanked turn would be driven by this effect more than the sideways component of engine thrust.


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What are the ICAO GNSS waypoint naming conventions?
I'm curious about ICAO naming conventions regarding GNSS and RNAV waypoints. Would I be correct in assuming that all GNSS area waypoints are 5 letters, and RNAV waypoints are 6?
What about waypoints like CB401? Do these represent a reference to a VOR/DME? If so, why are they included in GPS waypoint databases?

There are no "GNSS waypoints". GNSS is a common position source for RNAV, though not the only one.
All RNAV waypoints, like conventional (e.g. VOR/DME) fixes, have an identifier composed of five alphanumeric characters that is unique within a country. Some countries don't use numbers (e.g. the US); others do.


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How will a variable pitch propeller move during takeoff?
In my ATPL books i've learned that for a variable pitch propeller, it is better to have a fine pitch for take-off (higher RPM) and a coarse pitch for cruise (lower RPM).
In this question taken from the ATPL question bank it states that a propeller will move towards the coarse pitch on take off.
Does anybody understand why the correct answer for this question is coarse pitch. Is this implying that we initially start our take-off with a fine pitch but slowly change towards a coarse pitch during the climb?

What happens is the prop at idle will be at the maximum fine pitch limit.  The prop control is set to Max RPM for take-off (it uses variation in pitch to regulate RPM) so the prop governor is set up to regulate engine rpm at redline once there is sufficient power to spin it that fast.
You add throttle to take off.  The engine increases in speed, with the prop blades still in full fine pitch, until it gets to redline RPM, which is also the governor setting.  When the RPM gets there, the governor now acts to keep the RPM from going any higher.  It starts to move the blades to make the blade pitch more coarse to add load to the engine, and does so until the RPM stabilizes at the Max RPM governor setting.  As you keep adding throttle, increasing torque, the governor will increase blade angle more, whatever is required to keep the RPM at the governor setting.  Also, as you start to move, the forward speed starts to unload the propeller by reducing angle of attack, and the blades will coarsen further as required.
The governor doesn't know or care what the blade angle is; it's just a flyweight device like on an old steam engine and all it knows is "speed too high, do this, speed too low, do that". All the prop control does is apply a preload to a spring that determines the speed at which the flyweights will do their thing.
So yes the propeller will move toward coarse pitch because it's already fully fine and the governor has to move it more coarse to limit RPM in that situation.
If you shove the throttle in quickly, causing the RPM to rise very quickly, there may be a slight lag in the governor's response and the prop will overspeed momentarily a little bit (enough to create a noticeable EAAAOOOOWWW sound as RPM passes redline a bit and then slows down as it recovers to the governor setting).
The question is poorly worded.  It should be referring to "constant speed" governed propellers, not variable pitch, which is manually operated and quite rare (way too much work).  And the statement "best angle of attack" really means "angle of attack required to load the engine such that RPM is stablized at the Max RPM setting" (normally redline RPM for takeoff).  And this will always be more coarse from full fine pitch.

As airspeed increases, it gets easier for the prop to spin, which would cause a fixed-pitch prop to speed up.
A constant-speed prop's governor, however, will alter the prop pitch to be coarser (which makes it harder to spin) to counteract that and maintain the engine speed set by the prop lever.
An older variable-pitch (not constant-speed) prop requires you to adjust the prop lever manually as airspeed changes to get the same effect. However, such planes are rare these days.


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How does humidity's effects on density altitude get factored into Takeoff Distance Performance Charts?
Given that density altitude is a function of Pressure, Temperature, and Humidity, and most takeoff performance charts adjust takeoff distance as a function of pressure altitude and temperature, how is the humidity effect on density altitude taken into account?

As a comment rather than an answer because my experience is so dated (retired in 1999). Back then the answer to your question was that it wasn't taken into account. Our 747 performance charts had no available entry point for differing humidity, and that did occasionally cause concern. Go to https://terryliittschwager.com/talking-of-flying.php#dhahran. Then search the page for "very high humidity" to read of a takeoff at Dhahran, Saudi Arabia with a temp of 50C and humidity above 95%. I have no idea whether modern automatic systems have input for the humidity.

@Terry  at 50C and 95% RH around 10% of the air is water vapor (at 100C we have steam).

@RobertDiGiovanni Good to know. I wish I had known such in the later 1990s!

Temperature and humidity both affect air density and engine power. $H_2O$ molecules are lighter than the $N_2$ and $O_2$ molecules they displace in a volume of humid air.
1. Air Density. The graph below from this site shows that at 40°C, which is a high atmospheric temperature, dry air is about 0.94 density at 20°C, while 100% humidity reduces the density to about 0.91, a 3% reduction. The Standard Atmosphere references temps of 15°C, so the relative density reductions would be 0.92 for dry air and 0.89 for 100% moist air.

2. Engine power. Again from prof. Wittenberg's 50-year old uni book, the effects of humidity and temperature on engine power. Note that particularly piston engines are affected by humidity, turbojets & -fans not so much because they operate with a large over-ratio of air. Temperature affects a turbojet engine much more than a piston engine - the graph line is for a pure turbojet though, a modern high bypass fan engine graph would be much closer to the piston graph.


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Do ETOPS rated airplanes ALWAYS fly with ETOPS requirements?
So, as this is kind of a confusing subject I want to make sure I get a real answer, that's why I ended up doing it here. As ETOPS rated airplanes need to fly with certain rules so they comply with all the requirments, I am having kind of a hard time understanding if ETOPS rated airplanes need to comply with , for example, fuel requirements in a short flight that doesn't actually need ETOPS procedures. Example:
An A330 ETOPS rated aircraft is making a short flight from KPDX to KSFO, does it have to carry the extra fuel for ETOPS requierments, departing and landing with the extra weight and reducing performance?

The short answer is No (as per the regulations)
The ETOPS regulations state that an aircraft conducting ETOPS operations must be compliant to ETOPS regulations regarding its configuration (based on engine count). But aircraft capable of operating ETOPS flights only need to comply with the Advisory Circular when actually flying ETOPS. In other words the ETOPS AC only applies when the planed route falls under the guidance of the AC.

All two-engine airplanes and three- and four-engine passenger-carrying
  airplanes operated under part 121 are required to comply with §
  121.161. This regulation imposes special requirements for ETOPS for these airplanes. These operations are defined as:
(1) Two-Engine Airplanes. These are flights whose planned routing
  contains a point farther than 60 minutes flying time from an adequate
  airport at an approved one-engine inoperative cruise speed under
  standard conditions in still air.
(2) Passenger-Carrying Airplanes with More Than Two Engines. These are
  flights whose planned routing contains a point farther than 180
  minutes flying time from an adequate airport at an approved one-engine
  inoperative cruise speed under standard conditions in still air.

However, how an airline choses to route and use its fleet is up to them. Since ETOPS airframes may require different maintenance or other related things it may not be effective for a given airline to use ETOPS maintained aircraft on Non-ETOPS routes.

The airlines I worked for usually maintained ETOPS aircraft to ETOPS standards (requiring special checks and procedures) even when on short flights, but did not necessarily operate each flight under ETOPS rules (which would require the extra fuel etc.) - technically, it’s very hard to fit an ETOPS segment into a flight of less than two hours’ duration, anyway.


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Does a turboshaft's exhaust provide any thrust?
See the below figure. Does the exhaust gas coming out through the nozzle have any force?
If the exhaust gas has force then how is it controlled by the helicopter?

Welcome to the site. As your equation is stated now it is not clear to us what you are asking. Managed exhaust gas in what way? Which opposite forces does it not affect?

Please edit your question so that it is clear for people who first read it, you'll find an "edit" button underneath the question. I still don't understand what you're asking about opposite forces.

Are you asking about how a turbine engine works (how does the gas know to turn the turbine instead of going back in the compressor) or whether a helicopter turboshaft also generates thrust? Or something else?

Do you mean the rotor will push the exhaust back down in to the engine?

related: Can a gas turbine be able to switch from torque to thrust?

Yes, the exhaust of any engine does provide some thrust (very little for piston engines).
The amount of jet thrust for turboprops is in the range of 4%-15% of propeller-produced thrust. For a helicopter, it's only 2%-4% of rotor lift, but it's directed backwards. The helicopter in your picture would get 100-150 lbf of thrust from the engine exhaust, compared to 3,000-5,000 lbf of lift provided by the rotor.
This thrust is not controlled in any way. It helps a little in forward flight, and is compensated for by cyclic pitch in hover or maneuvering.
Residual thrust can be calculated, but is rarely specified in most sources. You can find some analyses of residual thrust for common engines online, such as for the PW-100 series, or for the Allison T-56.
The T-56's residual thrust, per above analysis, is 750 lbf, with 3442 kW on the shaft. That allows the T-56 engine to produce up to 10,000 lbf of thrust when used as a turboprop with a propeller. If used in a helicopter, such an engine would be able to deliver 25,000-40,000 lbf of lift with a rotor.

Awesome to know!  While I certainly don't doubt your claims, it would be great if you could add some sources to back them up.

If their is any source send me plz.

@dalearn Added a couple sources. Hard to find them online for helicopter engines specifically.

@Therac That is because the residual thrust of a turboshaft is really inconsequential, comparable to that of piston engines.

Indeed, for turboprops, the 'ideal' residual thrust for cruise is commensurate with the prop efficiency (for 85% prop, it would be about 15%), and the engines/nozzles are designed that way. For helicopter turboshafts, every bit of energy is better delivered to the shaft, and their nozzles are often designed with overexpansion (delivering lower than ambient pressure at the exhaust), which results in no significant thrust. With two engines, the nozzles are often directed sideways rather than backwards.


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What happens to thrust and drag in a straight and level flight?
Straight and level flight means when an aircraft is flying at a constant speed without changing its attitude. Its lift is equal to the weight of the aircraft. It is flying at a constant altitude.
Then I don't know what happens to thrust and drag?
Plz explain me

I assume here that the velocity is held constant.  For this, both thrust and drag must have forces of equal magnitude and opposite direction.  In short, they just balance each other out.

Actually not really - all 4 forces lift, weight, drag, and thrust in a sum balance out. Unfortunately, they are not always orthogonal, as the plane will quite certainly not fly with the axis of thrust 100% in line with the flight path = drag. But this is only a minute difference.

You got a proper answer. What do you not understand? (Please don't write in text speak.)

@Santhosh Welcome to Stack Exchange. I think it's rude to say "please give me proper answers" without explaining what additional information you're looking for. If you don't like this answer, please explain what you don't like about it.

If thrust is greater than drag, the aircraft accelerates. If thrust is less than drag, the aircraft decelerates. If thrust equals drag, then the aircraft remains at constant velocity. As commented by tsg, this ignores the fact that lift and drag are defined relative to the aircraft's path, while thrust is defined relative to the aircraft's orientation (pitch).


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Best range speed
I understand that the best L/D gives best range however, I have read that Carson speed is the most economical cruising speed where the ratio fuel flow/knot of airspeed reaches a minimum. I have read on a site that an increase in 16% in fuel flow will give a about a 31% increase in my speed.
My question is why this speed doesn’t give a better range than best range speed
Hypothetically, if I increase fuel flow by 20% and this gives me 20% increase in speed this should give me the same range as best range speed right! So if a 16% increase in fuel flow gives a 31% increase in speed this should give a better range in spite of a slight increase in drag or have I got this all wrong.

These number pairs you’re quoting are at specific points on a specific curve. They are not valid at any other point.

That site is wrong. The fuel flow (mass per time, proportional to power) will increase by more than 50%, but the total fuel used for flying a given distance will only increase by 16%, because you take ~30% less time to cover that distance.

Ok. This is my question. If total fuel used for a given distance will only increase by 16%  for a 30 knot increase in airspeed why don’t I get a greater range at Carson speed than at best range speed.


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Is there a blue book for private jets that state a price for sale if they have no documentation or logs?
I was wondering if there is an equivalent to the blue book of cars for planes that can inform how much a private jet can be sold or bought for if it doesn't have any documentation (for example, no maintenance books, no logs of any kind, no manual) but it's otherwise fully operational? For example how much is the sale price reduced in percentage terms compared to a jet with full documentation?

I'm not sure there is anything like this, it depends on how much of the log books can be reconstructed, and what checks/overhauls need to be performed after the effort. An aircraft without any logs is a significant cost to get airworthy, and that cost depends on a lot of factors.

You're more likely to find a Blue Book for aircraft with full documentation; having a multi-million $ asset with all documentation lost will be quite rare, since such a loss greatly reduces the value & utility of the aircraft.

@RalphJ Sometimes drug traffickers use and abandon private jets. I was wondering, how much sale value in terms of percentage does a private jet with no documentation lose?

@freethinker36 That depends on what percentage of the records can be reconstructed from various sources.

There is no "blue book" for aircraft without documentation. Your best bet is to research government and other auction sites that focus on the disposal of repossessed or confiscated aircraft and look at their historical sale records. Here is an example of the type of website you can check. But even with this info you will only get a vague idea of value, because there can be many other factors playing into the final value of such things. This is the kind of niche market where you can easily lose a lot of money if you don't know what you are doing, or make a lot of money if you do.

The price for a private jet with absolutely no documentation is, and I'm serious here, approximately the scrap value of metal in the plane. Some parts of the plane that are not subject to time or cycle based inspection and maintenance may be of some value depending on their condition.
If no records can be compiled, and no history established, the process of getting the plane certified back into operational status is going to be such a pita, it most likely is not worth the effort.
The possibility that such a plane would emerge is quite small. Unless the plane is reduced to a heap of scrap you'll have serial numbers all over the plane, so at least a part of the history can be tracked down. The longer the period of obscurity for the plane is, the deeper in certain matter a person wishing to operate the plane in a legal manner will be.

Context: In Guatemala, authorities discovered an abandoned private jet that had entered airspace the day before with drugs. They ordered its destruction the same day it was found, because they argued it was not deemed safe to fly and said they were following FAA regulations.

@freethinker36 most likely true reason was, that the plane was not likely to ever enter legal operations again for reasons I mentioned, and if the plane was stored, even for short periods of time, there is a high possibility it would end up in the same business it was doing before, trafficing drugs. Officials were being polite and discrete in their statement, not wanting to suggest a colleague of theirs might "bootleg" the plane  to criminals. Destroying it also sends a message that you lose stuff you use for criminal purposes. Although losing a 2million jet for 50 million profit, well...

Oh, and it just dawned on me: if you destroy the plane, like completely, you destroy evidence. As I wrote in my answer, the plane's history is easily recovered by a multitude of serial numebers etc, so the officials probably did not wan't that info to get spread out, for whatever reason...

A newspaper said a Guatemalan congressman was in that plane, which is why authorities wanted it destroyed asap. Just imagine the state of corruption Guatemala is in...


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Interpreting Missed Approach Section in Jeppesen Chart
While interpreting an RKPS(airport located in South Korea) LOC DME 06L, I found it somewhat confusing.

As you see, the initial action of the missed approach is climbing via heading 063 to GIDOO.
I can understand without any doubt if the chart states that I have to maintain track 063 rather than heading 063. I'm flying in C172, so the wind correction angle for maintaining track 063 will be big if the wind exists. But, if I maintain heading 063 as defined in the chart, I will not be able to reach GIDOO which might be determined by DME distance plus Radial from VOR or Localizer.
I tried to find GIDOO in other charts, but I couldn't.
So, what I want to know is, what should I maintain while executing missed approach procedure?
As mentioned in the chart, maintaining heading? or maintaining track?
In my opinion, we can't get any radial information from the localizer(we can't get a reliable signal from the course other than preselected final approach course), so I have to maintain heading 063 then when I reach 2 DME from ISAM, turn right heading 130.
Is it the correct action? or Do I need to maintain track 063? or.. Tracking back course localizer to GIDOO?

It is true, a track will send you directly over GIDOO but the missed approach text says to fly a heading.  It would take a strong wind to have any real effect on the offset from GIDOO when you cross it.
Most likely there are other aircraft on radar vectors and are being affected by the wind just like you would be so separation from an ATC standpoint is achieved.

So... you mean I have to maintain heading NOT track?
And if the separation is needed, the ATC will vector me to separate?

Yes on both questions.

thanks for your answer :)

I will need a little bit more context and need to see the entire approach plate, but it appears GIDOO is defined 2 NM DME from the localizer antenna along the localizer back course (this assumes the final approach heading is 063° magnetic as well).  We would need at least two navaids to locate and identify GIDOO.  This can be accomplished with the localizer DME, either along with localizer back course, or along a radial of the Saechon VORTAC which intersects GIDOO at 2 DME ISAM.


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When to use feeder route
The following section is from Instrument Procedures Handbook (FAA):

If a feeder route to an IAF begins at a fix located along the route of
flight prior to reaching the holding fix, and clearance for an
approach is issued, a pilot should commence the approach via the
published feeder route. For example, the aircraft would not be
expected to overfly the feeder route and return to it. The pilot is
expected to commence the approach in a similar manner at the IAF, if
the IAF for the procedure is located along the route of flight to the
holding fix.
If a route of flight directly to the IAF is desired, it should be so
stated by the controller with phraseology to include the words
“direct,” “proceed direct,” or a similar phrase that the pilot can
interpret without question. When a pilot is uncertain of the
clearance, ATC should be queried immediately as to what route of
flight is preferred.

Can you give examples to clarify it  a little bit more for each case. Pictures are appreciated.

I'm not sure I understand what you're asking. I suggest you look for an approach plate that's a good example of what you want to know, and try to ask something more specific using the details from that approach.

I found your quoted text in Section 5-4-6 of the AIM. You quote paragraphs b and c of the section. I will break down paragraphs a, b, and c using examples. Each paragraph is discussing a situation in which a pilot has been cleared to a holding fix (perhaps by clearance delivery, i.e. the last fix in the flight plan) and has been "cleared ... approach" prior to reaching that fix.
Paragraph 5-4-6 a

An aircraft which has been cleared to a holding fix and subsequently “cleared . . . approach” has not received new routing. Even though clearance for the approach may have been issued prior to the aircraft reaching the holding fix, ATC would expect the pilot to proceed via the holding fix (his/her last assigned route), and the feeder route associated with that fix (if a feeder route is published on the approach chart) to the initial approach fix (IAF) to commence the approach.

Summary: What to do when there is a feeder route from the holding fix to the IAF.
For example, prior to departure, a pilot calls clearance delivery and receives clearance to KMOB via V198 BROOKLEY VOR. They plan to shoot the ILS 15. Here is the approach plate and the chart.

Notice the feeder route from BROOKLEY to the IAF. If the pilot is "cleared ... approach" while on V198 transiting to BROOKLEY, their route has not changed. They must proceed to BROOKLEY as cleared, and then use the feeder from BROOKLEY to CAYAT (IAF). They may not turn directly to CAYAT unless specifically cleared to do so.

Paragraph 5-4-6 b

If a feeder route to an IAF begins at a fix located along the route of flight prior to reaching the holding fix, and clearance for an approach is issued, a pilot should commence the approach via the published feeder route; i.e., the aircraft would not be expected to overfly the feeder route and return to it. The pilot is expected to commence the approach in a similar manner at the IAF, if the IAF for the procedure is located along the route of flight to the holding fix.

Discusses: What to do when there is a feeder route from a fix on the route of flight prior to the holding fix.
In this example, a pilot plans to fly northbound to KMGM. They call clearance delivery and receive clearance to KMGM via V222 MONTGOMERY VOR. They plan to shoot the ILS 10.

Notice there is a feeder route from ALOON. If the pilot is on V222 heading north toward ALOON when they receive clearance for the approach, they will be expected to use the feeder from ALOON instead of overflying ALOON and proceeding all the way to MONTGOMERY VORTAC. In this case there is a feeder route from the holding fix as well, but that may not always be the case.

Paragraph 5-4-6 c

If a route of flight directly to the initial approach fix is desired, it should be so stated by the controller with phraseology to include the words “direct . . .,” “proceed direct” or a similar phrase which the pilot can interpret without question. When uncertain of the clearance, immediately query ATC as to what route of flight is desired.

Summary: If a pilot would like to fly directly to an IAF without using a feeder route, they can request to do so. The clearance must contain the specified phraseology.

Exactly what I was looking for. Awesome. Thank you.